Year 11 Physics Module 4 ⏱ ~35 min 5 MC · 3 Short Answer Lesson 6 of 17

Series Circuits

General Electric Christmas tree lights (US patent 1917): 25 incandescent bulbs wired in series; total supply 110 V; each bulb received 110/25 = 4.4 V. If one bulb's filament broke, total resistance → ∞, current = 0, and all lights went out simultaneously. The series-circuit "one out, all out" problem was so frustrating that parallel Christmas lights replaced series strings by the 1950s.

Today's hook: General Electric's US patent 1917 covered a 25-bulb Christmas tree light string wired entirely in series across 110 V, each bulb received 110 ÷ 25 = 4.4 V. When one filament broke, total resistance jumped to infinity, current dropped to zero, and every bulb went dark simultaneously. The "one out, all out" failure mode was so commercially disastrous that parallel wiring replaced series strings by the 1950s. Understanding exactly why requires the three rules of series circuits.
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Warm up first

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Before you read, predict

Three identical light globes (each 6 Ω) are connected in series to a 12 V battery.

Predict 1: What happens to all three globes if one globe's filament breaks open?

Predict 2: If each globe is identical, how does the voltage shared across each globe compare to the supply voltage?

In a series circuit with two resistors, the current through each resistor is:

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Learning Intentions

Learning Intentions

Know

  • Series circuit rules: same current, voltages add, resistances add
  • $R_{\text{total}} = R_1 + R_2 + R_3 + \ldots$
  • $V_{\text{supply}} = V_1 + V_2 + V_3 + \ldots$
  • $I$ is the same at every point in a series circuit

Understand

  • Why current is the same everywhere (conservation of charge)
  • Why voltages add to the supply (conservation of energy)
  • How the voltage divides in proportion to resistance

Can Do

  • Calculate total resistance in a series circuit
  • Find current, voltage and resistance for each component
  • Predict the effect of adding or removing a resistor
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Key Terms

Key Terms
Series circuitA circuit with only one path for current to flow. All components are connected end-to-end.
Voltage dividerIn a series circuit, the voltage divides across components in proportion to their resistance: $V_1/V_2 = R_1/R_2$.
Conservation of chargeIn a series circuit, the same current flows everywhere because charge cannot accumulate at any point.
Conservation of energyVoltages across components add to the supply voltage, all energy given to charges by the battery is transferred to components.
Cross-lesson links: L05 defined resistance, current, and voltage. L06 applies these to series circuits, where components share the same current. Key series rules: same current everywhere; voltages add to supply; total resistance = sum of individual resistances. These rules derive directly from conservation of energy and conservation of charge.
Misconceptions to fix
✗ Wrong: In a series circuit, the current gets "used up" as it passes through each resistor.
✓ Right: Current (charge per second) is the same everywhere in a series circuit, charge cannot be created or destroyed at any junction. It is voltage (energy per charge) that gets used up, not current.
✗ Wrong: Adding more resistors to a series circuit increases the current.
✓ Right: Adding resistors in series increases total resistance. With the same supply voltage, $I = V/R_{\text{total}}$ decreases.
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Series Circuit Rules

Core Content

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Series Circuit Rules
+5 XP

Take a battery, three small globes, and some wire. Connect the globes one after the other in a single loop with the battery, all three glow. Now unscrew any one globe from its socket and all three go out instantly, even though the other two are physically intact. Slide the unscrewed globe back in and all three light up simultaneously. This is the visual signature of a series circuit: one path, so breaking it anywhere kills the whole circuit. There are three rules that follow directly from this single-path structure.

Series circuit rules

1. Same current: $I_1 = I_2 = I_3 = \ldots = I_{\text{supply}}$

2. Voltages add: $V_{\text{supply}} = V_1 + V_2 + V_3 + \ldots$

3. Resistances add: $R_{\text{total}} = R_1 + R_2 + R_3 + \ldots$

Voltage Divider Rule

Because the current is the same through all components, the voltage across each component is proportional to its resistance:

Voltage divider

$V_1 = I R_1 = V_{\text{supply}} \times \dfrac{R_1}{R_{\text{total}}}$

Worked example, three resistors in series

$R_1 = 10\ \Omega$, $R_2 = 20\ \Omega$, $R_3 = 30\ \Omega$ connected in series to a 12 V supply. Find $I$, $V_1$, $V_2$, $V_3$.

  1. $R_{\text{total}} = 10 + 20 + 30 = 60\ \Omega$
  2. $I = V/R_{\text{total}} = 12/60 = 0.20\ \text{A}$
  3. $V_1 = IR_1 = 0.20 \times 10 = 2.0\ \text{V}$
  4. $V_2 = IR_2 = 0.20 \times 20 = 4.0\ \text{V}$
  5. $V_3 = IR_3 = 0.20 \times 30 = 6.0\ \text{V}$
  6. Check: $V_1 + V_2 + V_3 = 2.0 + 4.0 + 6.0 = 12\ \text{V}$ ✓

In a series circuit: current is identical through all components ($I_1 = I_2 = \ldots$); supply voltage equals the sum of component voltages; total resistance $R_{\text{total}} = R_1 + R_2 + \ldots$; voltage divides in proportion to resistance.

Pause, copy the highlighted rules into your book before moving on.

In a series circuit, the current through each component is the same.

Adding a resistor in series increases the total current in the circuit.

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Activity 1, Series Circuit Calculations

Activity 1, Series Circuit Calculations
ApplyBand 3

Three resistors, $R_1 = 15\ \Omega$, $R_2 = 25\ \Omega$, $R_3 = 10\ \Omega$, are connected in series to a 30 V battery.

  1. Calculate the total resistance of the circuit.
  2. Calculate the current in the circuit.
  3. Calculate the voltage across each resistor and verify they add to the supply voltage.

In a series circuit with $R_1 = 4\ \Omega$ and $R_2 = 12\ \Omega$ connected to a 24 V supply, the voltage across $R_1$ is:

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Activity 2, Voltage Divider Analysis

Activity 2, Voltage Divider Analysis
AnalyseBand 4

Two resistors, $R_A = 40\ \Omega$ and $R_B = 60\ \Omega$, are connected in series to a 10 V supply.

  1. Calculate the voltage across each resistor.
  2. Show that $V_A/V_B = R_A/R_B$.
  3. If $R_A$ is replaced by a 20 Ω resistor, what happens to the voltage across $R_B$? Calculate the new $V_B$.

Three of these are correct series circuit rules. Pick the odd one out.

Three resistors of 5 Ω, 10 Ω, and 15 Ω are connected in series. The total resistance is _____ Ω.

In a series circuit, the component with the largest resistance has the:

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Wrap-up, Series Circuit Summary

Wrap-up, Series Circuit Summary

Current

$I_1 = I_2 = I_3$ everywhere. Conservation of charge: charge cannot accumulate.

Voltage

$V_{\text{supply}} = V_1 + V_2 + V_3$. Conservation of energy: all EMF transferred to components.

Resistance

$R_{\text{total}} = R_1 + R_2 + R_3$. Adding components always increases total resistance.

Voltage divider

$V_n = I R_n = V_{\text{supply}} \times R_n/R_{\text{total}}$. Large $R$ → large $V$ drop.

✓

Show what you have learned

Quick recall, series circuits
+5 XP
Short Answer, 10 marks
+5 XP

ApplyBand 3(3 marks) 3. Three resistors, 6 Ω, 9 Ω, and 15 Ω, are connected in series to a 24 V battery. Calculate (a) the total resistance, (b) the current in the circuit, and (c) the voltage across the 9 Ω resistor.

AnalyseBand 4(3 marks) 4. In a series circuit with $R_A = 30\ \Omega$ and $R_B = 70\ \Omega$ connected to a 20 V battery: (a) Calculate the ratio $V_A : V_B$. (b) Explain why this ratio equals $R_A : R_B$. (c) If $R_A$ is doubled, what happens to $V_B$?

EvaluateBand 6(4 marks) 5. A shopkeeper proposes connecting all the shop's display lights (each 6 Ω) in series from a 240 V mains supply. Evaluate this proposal with reference to series circuit behaviour, and explain why a parallel connection would be preferred.

Show all answers

Activity 1, Model Answers

  1. $R_{\text{total}} = 15 + 25 + 10 = 50\ \Omega$
  2. $I = 30/50 = 0.60\ \text{A}$
  3. $V_1 = 0.60 \times 15 = 9.0\ \text{V}$; $V_2 = 0.60 \times 25 = 15\ \text{V}$; $V_3 = 0.60 \times 10 = 6.0\ \text{V}$. Sum = 30 V ✓

Activity 2, Model Answers

  1. $R_{\text{total}} = 100\ \Omega$; $I = 10/100 = 0.10\ \text{A}$. $V_A = 0.10 \times 40 = 4.0\ \text{V}$; $V_B = 0.10 \times 60 = 6.0\ \text{V}$.
  2. $V_A/V_B = 4/6 = 2/3$; $R_A/R_B = 40/60 = 2/3$ ✓.
  3. New $R_{\text{total}} = 20 + 60 = 80\ \Omega$; $I = 10/80 = 0.125\ \text{A}$; new $V_B = 0.125 \times 60 = 7.5\ \text{V}$. $V_B$ increases when $R_A$ decreases.

Short Answer, Model Answers

Q3 (3 marks): (a) $R_{\text{total}} = 6 + 9 + 15 = 30\ \Omega$. (b) $I = 24/30 = 0.80\ \text{A}$. (c) $V_9 = 0.80 \times 9 = 7.2\ \text{V}$.

Q4 (3 marks): (a) $I = 20/100 = 0.20\ \text{A}$; $V_A = 6.0\ \text{V}$; $V_B = 14\ \text{V}$; ratio $6:14 = 3:7$; $R_A:R_B = 30:70 = 3:7$ ✓. (b) In series, $I$ is the same; $V = IR$, so voltage ratio = resistance ratio. (c) If $R_A$ doubles to 60 Ω: $R_{\text{total}} = 130\ \Omega$; $I = 20/130 = 0.154\ \text{A}$; $V_B = 0.154 \times 70 = 10.8\ \text{V}$ (increases slightly).

Q5 (4 marks): In series: if one globe fails (open-circuit), all others go out, not practical for display lighting. The voltage divides across the globes, so each receives only 240/n V (where n = number of globes), not the full 240 V, they would each dim as more globes are added. In parallel, each globe receives the full 240 V and operates at designed brightness; one failing does not affect others; globes can be switched on/off independently. Parallel connection is far preferable for commercial display lighting.

✓

Retrieve, reflect and finish

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How did your thinking change?

The General Electric Christmas tree light string from US patent 1917 wired 25 bulbs in series across 110 V, 4.4 V per bulb. When one filament failed, the circuit was broken and all 25 bulbs went out simultaneously, because there was only one current path. That is the defining feature of a series circuit: break any single link and the whole chain dies.

Now look back at your Think First answers: one globe failing breaks the single path, so all three go out, exactly like those 1917 lights. Three equal 6 Ω globes on 12 V each receive 12 ÷ 3 = 4 V (voltage divides equally when resistances are equal).

← Current, Voltage & Resistance Next: Parallel Circuits →

Module overview · Physics · Checkpoint 1 · Module quiz